Question

If A, B, C are the angles of a triangle, then
$\sin \left(\frac{B+C}{2}\right)=$ ___.

• A. $\cot \frac{A}{2}$
• B. $\cos \frac{A}{2}$
• C. $\sin \frac{A}{2}$
• D. $\tan \frac{A}{2}$

Answer 1

Answer: (B)

$\cos \frac{A}{2}$

$A+B+C={180}^{\circ }$
(sum of angles of a triangle)

$⟹B+C={180}^{\circ }-A$

$\therefore \frac{B+C}{2}=\frac{{180}^{\circ }-A}{2}$

$\frac{B+C}{2}={90}^{\circ }-\frac{A}{2}$

Taking sine on both sides we get,

$\sin \left(\frac{B+C}{2}\right)=\sin ({90}^{\circ }-\frac{A}{2})$

$\therefore \sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$